[/quote]
,
inscribed in 
with 
being the midpoint of 
.
of 
.
be the projection of 
onto 
.
that is not on 
intersects 
at 
.
and 
intersect at 
.
is an isosceles triangle.
be the set of all positive integers. We say that a function 
is Georgian if 
and, for every positive integer 
,
such that![\[
f^{(k)}(n) = 1, \quad \text{where } f^{(k)} = f \circ f \cdots \circ f \quad \text{(applied } k \text{ times)}.
\]](http://latex.artofproblemsolving.com/5/3/c/53ca7c663237d939a7a049aa3569f8d595b259cf.png)
If 
is a Georgian function, we define, for each positive integer 
as the smallest positive integer 
such that 
.
for which there exists a Georgian function such that, for every positive integer 
,
.
prove that
denotes the sum of the integers, 
denotes the sum of the square of the integers and 
denotes the sum of cubes of the integers(that is, if the three integers are 
,
,
,
)
and 
,
.
and 
that satisfy the system of equations:
and 
be the sets of integers and rationals respectively.
such that the sets 
are disjoint?
denotes the set 
,
and for 
.
be positive real numbers such that 
.![\[ \frac{x^2-yz}{x^2+x}+\frac{y^2-zx}{y^2+y}+\frac{z^2-xy}{z^2+z}\leq 0. \]](http://latex.artofproblemsolving.com/e/a/1/ea1f56cc594862673040411c3ad322cb00cfe904.png)
grid 
)
black cells.Detemine the number of balanced colorings.
be a quadrilateral inscribed in a circle with center 
and 
.
be the circumcircle of 
and 
be the circumcircle of 
.
.
.
.
denote the number of 
s![\[\sum_{i=1}^{n}(-1)^{s(3i)} > 0.\]](http://latex.artofproblemsolving.com/d/e/9/de975cc2632685d7a3f71c023115535d0aeb5dbb.png)
Holden Mui
for which there exist real numbers 
satisfying 
,
and
for 
.
be an integer. Find the smallest integer 
with the property that there exists a set of 
-
Y by samrocksnature, Adventure10, ImSh95, Mango247, and 1 other user
Let 
be an increasing sequence of nonnegative integers such that every nonnegative integer can be expressed uniquely in the form 
, where 
and 
are not necessarily distinct. Determine 
.
be an increasing sequence of nonnegative integers such that every nonnegative integer can be expressed uniquely in the form 
, where 
and 
are not necessarily distinct. Determine 
.Attachments:
This post has been edited 1 time. Last edited by orl, Oct 23, 2004, 12:48 PM
.
can be represented in the mentioned way, we must have 
(this will be important a bit later). We have 
.
and we get 
.
.
,
,
we have 
,
,
.
'
,
.
.
is the sum of the first 
numbers. So 
.
.
is the sum of the first 
is simply the result of writing 
and adding it on to the current sequence.
for some integer 
.
.
,
has exactly 
,
equals 
,
in the desired form.
digits, possibly with leading 0s, so we have 
.
in base 8, 
in base 8, and 
in base 8.
with 
,
possible numbers of the form 
and 
for any integer 
,
.
as 
and observe that this formal series is defined at all values 
of the variable. The given relation is equivalent to 
Replacing 
by 
reveals that 
and we conclude that the formal series 
is given by 
Thus, 
when written in base 
where 
,
.![\[a_n = \sum_{i\geq 0} e_i\cdot 8^{i}\]](http://latex.artofproblemsolving.com/8/1/1/811496f4572bcdc41c162211ddb8698e91595004.png)
,
,
.
,
![\[1998 = 11111001110_2 \Longrightarrow a_{1998}=11111001110_8 = 1227096648_{10}\]](http://latex.artofproblemsolving.com/a/2/f/a2f0971eac26d469933ffd73e7f02c425c2c1653.png)
and we are done.
contains precisely the nonnegative integers that contain no digits other than 0 and 1 in base 8. Clearly, this sequence works, since for each nonnegative integer, we essentialy express it in base 8, split each digit into 3 bits, and use those to pick either 0 or 1 in that corresponding location in order to find the bit in that place for 
.
,
denote the smallest nonnegative integer that cannot be expressed as 
.
.
,
as 
However, we can also express 
,
,
for all 
However, consider 
through 
for 
,
.
.
On the other hand, combining this with 
,
,
and thus 
can be characterized as the sequence of numbers expressible as the sum of powers of 
,
,
,
.
,
and the third being every digit of place value 
and the number formed only using digit first class is 
,
and the third class forms 
.
t
terms. We claim by induction that the first 
integers.
.
possibilities are expressed 
we simply need those 
up to 
.
.
.
so 
so 
.
is a term in 
,
is a term in 
,
must both be terms in 
,
.
and 
,
bits determine 
bits determine 
bits determine 
,
.
,![\[f(x)f(x^2)f(x^4) = 1 + x + x^2 + \dots = \dfrac{1}{1 - x} \implies \dfrac{f(x)}{f(x^8)} = \dfrac{f(x)f(x^2)f(x^4)}{f(x^2)f(x^4)f(x^8)} = \dfrac{1 - x^2}{1 - x} = 1 + x.\]](http://latex.artofproblemsolving.com/9/4/f/94f8f5bd31d28d7dbb6953d923ac0b401e973b98.png)
Therefore, we have the telescoping series![\[f(x) = \dfrac{f(x)}{f(x^8)} \cdot \dfrac{f(x^8)}{f(x^{8^2})} \cdot \dfrac{f(x^{8^2})}{f(x^{8^3})} \cdots = (1 + x)(1 + x^8)(1 + x^{8^2}) \cdots \]](http://latex.artofproblemsolving.com/7/4/d/74d164c9ffcff327b0696379c49cd65471b8eab1.png)
So ![\[1998 = 11111001110_2 \implies a_{1998} = 8^{10} + 8^9 + 8^8 + 8^7 + 8^6 + 8^3 + 8^2 + 8^1\]](http://latex.artofproblemsolving.com/4/2/7/427f6c438222c5a01596beefc19e669bfb242710.png)
as wanted.
and 
with 
.
(if not we can't get 1). As 
can be achieved by using 1 and 0, but not 8, 
and 
.
uniquely. Write 
,
and 
,
.
,
,
,
,
,
,
.
.
,
(because 
)
=
.
,
just using 
by induction hypothesis. Therefore, we select those exact indices for 
instead of 
.
just using 
,
,
for 
.
and as 
,
and 
as otherwise we would be unable to represent 
The following claim determines that at most one sequence works:
for 
If 
then we could represent the number 
then we can never form 
where 
is the binary representation of 
This works because the only way to account for the 
positions in the binary representation are with the 
and 
positions. This uniquely determines the 
Hence this is the only sequence which works. Finally, 
so our answer is 
and then eventually we reach a number not representable, in which we have to add that number to the sequence. This process is deterministic.
inclusive (uniquely) all numbers can be represented uniquely by considering each digit separately. Therefore 
Hence the answer is 
